# Computing Expectation of Normal and Log-Normal distributed variables

I need to calculate the variance of $$E = \sqrt {(X - a)^2 + (Y -b)^2} - Z ~,$$ where $$X$$ and $$Y$$ are Normally distributed variables with zero mean and $$Z$$ is Log-Normally distributed. All $$X, Y$$, and $$Z$$ are uncorrelated with known mean and variances, while $$a$$ and $$b$$ are constants.

I am very much new to this area but trying calculate variance of $$E$$. My questions are as follows:

1. Is the following correct? $$\mathrm{Var}(E) = \mathrm{Var}\left( \sqrt {(X - a)^2 + (Y -b)^2}\right ) + \mathrm{Var}(Z)~.$$ This identity is true for two independent Normal distributed variables but I doubt that this will be true for this case as well.
2. Can somebody point out some resource for calculating the variance of square root terms?
• To make a square root of a big expression, use \sqrt{expression} with curly braces. To make left and right parentheses than look nice, use \left( and \right). – 6005 Oct 24 '16 at 3:39

## 1 Answer

The answer to your first question is yes: Because $X$, $Y$, $Z$ are independent, then $W = ((X-a)^2 + (Y-b)^2)^{1/2}$ is independent of $Z$, hence $$\operatorname{Var}[E] = \operatorname{Var}[W] + \operatorname{Var}[Z].$$

The answer to your second question is harder. We can observe, however, $$\operatorname{Var}[W] = \operatorname{E}[W^2] - \operatorname{E}[W]^2,$$ and the expectation of $W^2$ is actually easy, because $(X-a)^2 + (Y-b)^2 \ge 0$ for all $X, Y, a, b$. Linearity of expectation gives us $$\operatorname{E}[W^2] = \operatorname{E}[(X-a)^2] + \operatorname{E}[(Y-b)^2],$$ and since $X$ is normal with mean $0$ (you did not specify the variances of $X$ and $Y$), we have $$\operatorname{E}[(X-a)^2] = \operatorname{E}[X^2] - 2a \operatorname{E}[X] + a^2 = \operatorname{Var}[X] + \operatorname{E}[X]^2 - 0 + a^2 = \sigma_X^2 + a^2,$$ where $\sigma_X^2$ is the variance of $X$. It follows that $$\operatorname{E}[W^2] = \sigma_X^2 + \sigma_Y^2 + a^2 + b^2.$$ The expectation of $W$ itself is more problematic; at this time, I don't have an elegant solution; I can only suggest directly evaluating the integral $$\operatorname{E}[W] = \int_{x=-\infty}^\infty \int_{y=-\infty}^\infty ((x-a)^2 + (y-b)^2)^{1/2} \frac{e^{-x^2/(2\sigma_X^2)} e^{-y^2/(2\sigma_Y^2)}}{2\pi \sigma_X \sigma_Y} \, dy \, dx.$$ But I suspect this integral does not have an elementary closed form. I doubt it is integrable even when $\sigma_X = \sigma_Y = 1$.

• Thanks for such a clear answer. i will evaluate integration part. – vicky Oct 24 '16 at 5:23
• yes! you are right this E[W] integral does not turn out be closed form . Is there any way to approximate this function which can give me a closed form of integral . – vicky Oct 24 '16 at 7:40
• thanks for comments. please give me any direction on the issue mentioned in above comment – vicky Oct 24 '16 at 7:45